Question

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Answer 1

Answer:

a) $x = -1$ and  $y = 5$, b) $x = 2$ and $y = -1$, c) $x = -2$ and $y = 1$, d) $x = 2$ and $y = -9$, e) $x = 2$ and $y = -2$, f) $x = 2$ and $y = -1$, g) $x = 2$ and $y = -1$, h) $x = 1$ and $y = 2$.

Step-by-step explanation:

Each system is solved as follows (Cada sistema es resuelto como sigue):

a) $2\cdot x + y = 3$ and $3\cdot x - y = -8$

$y = 3 - 2\cdot x$

$3\cdot x - 3 + 2\cdot x = -8$

$5\cdot x = -5$

$x = -1$

$y = 5$

b) $x - 2\cdot y = 4$ and $-x+3\cdot y = -5$

$x = 4 + 2\cdot y$

$-4-2\cdot y+3\cdot y = - 5$

$-4+y = -5$

$y = -1$

$x = 2$

c) $-2\cdot x + 5\cdot y = 9$ and $x - y = -3$

$x = y-3$

$-2\cdot y +6 +5\cdot y = 9$

$3\cdot y = 3$

$y = 1$

$x = -2$

d) $5\cdot x - 4\cdot y = 2$ and $3\cdot x + 2\cdot y = -12$

$3\cdot x + 2\cdot y = -12$

$6\cdot x + 4\cdot y = -24$

$4\cdot y = -6\cdot x -24$

$5\cdot x +6\cdot x +24 = 2$

$11\cdot x = -22$

$x = 2$

$y = -9$

e) $x + y = 0$ and $2\cdot x +3\cdot y = -2$

$y = -x$

$2\cdot x -3\cdot x = -2$

$-x = -2$

$x = 2$

$y = -2$

f) $3\cdot x + 5\cdot y = 1$ and $x + y = 1$

$y = 1 - x$

$3\cdot x + 5 - 5\cdot x = 1$

$-2\cdot x = -4$

$x = 2$

$y = -1$

g) $5\cdot x - 2\cdot y = 12$ and $4\cdot x + 3\cdot y = 5$

$x = \frac{12+2\cdot y}{5}$

$x = \frac{5-3\cdot y}{4}$

$\frac{12+2\cdot y}{5} = \frac{5-3\cdot y}{4}$

$4\cdot (12+2\cdot y) = 5\cdot (5-3\cdot y)$

$48 + 8\cdot y = 25 - 15\cdot y$

$23\cdot y = -23$

$y = -1$

$x = 2$

h) $7\cdot x + 8\cdot y = 23$ and $3\cdot x + 2\cdot y = 7$

$3\cdot x + 2\cdot y = 7$

$12\cdot x + 8\cdot y = 28$

$8\cdot y = 28 - 12\cdot x$

$7\cdot x +28-12\cdot x = 23$

$-5\cdot x = -5$

$x = 1$

$y = 2$